Question: $ \int_0^1 \int_0^{2y} dx \, dy + \int_1^2 \int_0^{2 - 2(y - 1)} dx \, dy$ Switch the bounds of the double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^2 \int_{x/4}^{1 - x/4} dx \, dy$ (Choice B) B $ \int_0^2 \int_{x/2}^{2 - x/2} dx \, dy$ (Choice C) C $ \int_1^2 \int_{x/2}^{2 - x} dx \, dy$ (Choice D) D $ \int_1^2 \int_{x/4}^{1 - x/4} dx \, dy$
Explanation: The first step whenever we want to switch bounds is to sketch the region of integration that we're given. We are given two regions in this case. The first has $0 < x < 2y$ and $0 < y < 1$. The second has $0 < x < 2 - 2(y - 1)$ and $1 < y < 2$. Therefore: ${1}$ ${2}$ ${3}$ ${1}$ ${2}$ ${3}$ $y$ $x$ Because we're switching bounds to $dy \, dx$, we need to start with numeric bounds for $x$. We see that $0 < x < 2$. Then we can define $y$ in terms of $x$. Thus, $\dfrac{x}{2} < y < 2 - \dfrac{x}{2}$. In conclusion, the double integral after switching bounds is: $ \int_0^2 \int_{x/2}^{2 - x/2} dx \, dy$